Find the Concavity natural log of x^2+9

$\ln (x^{2} + 9)$

Write $\ln (x^{2} + 9)$ as a function.

$f (x) = \ln (x^{2} + 9)$

Find the $x$ values where the second derivative is equal to $0$ .

Find the second derivative.

Find the first derivative.

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{2} + 9$ .

To apply the Chain Rule, set $u$ as $x^{2} + 9$ .

$f' (x) = \frac{d}{d u} (\ln (u)) \frac{d}{d x} (x^{2} + 9)$

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$f' (x) = \frac{1}{u} \cdot \frac{d}{d x} (x^{2} + 9)$

Replace all occurrences of $u$ with $x^{2} + 9$ .

$f' (x) = \frac{1}{x^{2} + 9} \cdot \frac{d}{d x} (x^{2} + 9)$

Differentiate.

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f' (x) = \frac{1}{x^{2} + 9} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = \frac{1}{x^{2} + 9} \cdot (2 x + \frac{d}{d x} (9))$

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{x^{2} + 9} \cdot (2 x + 0)$

Combine fractions.

Add $2 x$ and $0$ .

$f' (x) = \frac{1}{x^{2} + 9} \cdot (2 x)$

Combine $2$ and $\frac{1}{x^{2} + 9}$ .

$f' (x) = \frac{2}{x^{2} + 9} \cdot x$

Combine $\frac{2}{x^{2} + 9}$ and $x$ .

$f' (x) = \frac{2 x}{x^{2} + 9}$

Find the second derivative.

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2 x}{x^{2} + 9}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{x}{x^{2} + 9}]$ .

$f'' (x) = 2 \frac{d}{d x} (\frac{x}{x^{2} + 9})$

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x^{2} + 9$ .

$f'' (x) = 2 (\frac{(x^{2} + 9) \frac{d}{d x} (x) - x \frac{d}{d x} (x^{2} + 9)}{{(x^{2} + 9)}^{2}})$

Differentiate.

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 2 (\frac{(x^{2} + 9) \cdot 1 - x \frac{d}{d x} (x^{2} + 9)}{{(x^{2} + 9)}^{2}})$

Multiply $x^{2} + 9$ by $1$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - x \frac{d}{d x} (x^{2} + 9)}{{(x^{2} + 9)}^{2}})$

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - x (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))}{{(x^{2} + 9)}^{2}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - x (2 x + \frac{d}{d x} (9))}{{(x^{2} + 9)}^{2}})$

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - x (2 x + 0)}{{(x^{2} + 9)}^{2}})$

Simplify the expression.

Add $2 x$ and $0$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - x (2 x)}{{(x^{2} + 9)}^{2}})$

Multiply $2$ by $- 1$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - 2 x \cdot x}{{(x^{2} + 9)}^{2}})$

Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - 2 (x \cdot x)}{{(x^{2} + 9)}^{2}})$

Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - 2 (x \cdot x)}{{(x^{2} + 9)}^{2}})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 2 (\frac{x^{2} + 9 - 2 x^{1 + 1}}{{(x^{2} + 9)}^{2}})$

Add $1$ and $1$ .

$f'' (x) = 2 (\frac{x^{2} + 9 - 2 x^{2}}{{(x^{2} + 9)}^{2}})$

Subtract $2 x^{2}$ from $x^{2}$ .

$f'' (x) = 2 (\frac{- x^{2} + 9}{{(x^{2} + 9)}^{2}})$

Combine $2$ and $\frac{- x^{2} + 9}{{(x^{2} + 9)}^{2}}$ .

$f'' (x) = \frac{2 (- x^{2} + 9)}{{(x^{2} + 9)}^{2}}$

Simplify.

Apply the distributive property.

$f'' (x) = \frac{2 (- x^{2}) + 2 \cdot 9}{{(x^{2} + 9)}^{2}}$

Simplify each term.

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{- 2 x^{2} + 2 \cdot 9}{{(x^{2} + 9)}^{2}}$

Multiply $2$ by $9$ .

$f'' (x) = \frac{- 2 x^{2} + 18}{{(x^{2} + 9)}^{2}}$

The second derivative of $f (x)$ with respect to $x$ is $\frac{- 2 x^{2} + 18}{{(x^{2} + 9)}^{2}}$ .

$\frac{- 2 x^{2} + 18}{{(x^{2} + 9)}^{2}}$

Set the second derivative equal to $0$ then solve the equation $\frac{- 2 x^{2} + 18}{{(x^{2} + 9)}^{2}} = 0$ .

Set the second derivative equal to $0$ .

$\frac{- 2 x^{2} + 18}{{(x^{2} + 9)}^{2}} = 0$

Set the numerator equal to zero.

$- 2 x^{2} + 18 = 0$

Solve the equation for $x$ .

Subtract $18$ from both sides of the equation.

$- 2 x^{2} = - 18$

Divide each term in $- 2 x^{2} = - 18$ by $- 2$ and simplify.

Divide each term in $- 2 x^{2} = - 18$ by $- 2$ .

$\frac{- 2 x^{2}}{- 2} = \frac{- 18}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x^{2}}{- 2} = \frac{- 18}{- 2}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 18}{- 2}$

Simplify the right side.

Divide $- 18$ by $- 2$ .

$x^{2} = 9$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

Find the domain of $f (x) = \ln (x^{2} + 9)$ .

Set the argument in $\ln (x^{2} + 9)$ greater than $0$ to find where the expression is defined.

$x^{2} + 9 > 0$

Solve for $x$ .

Subtract $9$ from both sides of the inequality.

$x^{2} > - 9$

Since the left side has an even power, it is always positive for all real numbers.

All real numbers

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Substitute any number from the interval $(- \infty, - 3)$ into the second derivative and evaluate to determine the concavity.

Replace the variable $x$ with $- 6$ in the expression.

$f'' (- 6) = \frac{- 2 {(- 6)}^{2} + 18}{{({(- 6)}^{2} + 9)}^{2}}$

Simplify the result.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$f'' (- 6) = \frac{- 2 \cdot 36 + 18}{{({(- 6)}^{2} + 9)}^{2}}$

Multiply $- 2$ by $36$ .

$f'' (- 6) = \frac{- 72 + 18}{{({(- 6)}^{2} + 9)}^{2}}$

Add $- 72$ and $18$ .

$f'' (- 6) = \frac{- 54}{{({(- 6)}^{2} + 9)}^{2}}$

Simplify the denominator.

Raise $- 6$ to the power of $2$ .

$f'' (- 6) = \frac{- 54}{{(36 + 9)}^{2}}$

Add $36$ and $9$ .

$f'' (- 6) = \frac{- 54}{45^{2}}$

Raise $45$ to the power of $2$ .

$f'' (- 6) = \frac{- 54}{2025}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $- 54$ and $2025$ .

Factor $27$ out of $- 54$ .

$f'' (- 6) = \frac{27 (- 2)}{2025}$

Cancel the common factors.

Factor $27$ out of $2025$ .

$f'' (- 6) = \frac{27 \cdot - 2}{27 \cdot 75}$

Cancel the common factor.

$f'' (- 6) = \frac{27 \cdot - 2}{27 \cdot 75}$

Rewrite the expression.

$f'' (- 6) = \frac{- 2}{75}$

Move the negative in front of the fraction.

$f'' (- 6) = - \frac{2}{75}$

The final answer is $- \frac{2}{75}$ .

$- \frac{2}{75}$

The graph is concave down on the interval $(- \infty, - 3)$ because $f'' (- 6)$ is negative.

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Substitute any number from the interval $(- 3, 3)$ into the second derivative and evaluate to determine the concavity.

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = \frac{- 2 {(0)}^{2} + 18}{{({(0)}^{2} + 9)}^{2}}$

Simplify the result.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$f'' (0) = \frac{- 2 \cdot 0 + 18}{{(0^{2} + 9)}^{2}}$

Multiply $- 2$ by $0$ .

$f'' (0) = \frac{0 + 18}{{(0^{2} + 9)}^{2}}$

Add $0$ and $18$ .

$f'' (0) = \frac{18}{{(0^{2} + 9)}^{2}}$

Simplify the denominator.

Raising $0$ to any positive power yields $0$ .

$f'' (0) = \frac{18}{{(0 + 9)}^{2}}$

Add $0$ and $9$ .

$f'' (0) = \frac{18}{9^{2}}$

Raise $9$ to the power of $2$ .

$f'' (0) = \frac{18}{81}$

Cancel the common factor of $18$ and $81$ .

Factor $9$ out of $18$ .

$f'' (0) = \frac{9 (2)}{81}$

Cancel the common factors.

Factor $9$ out of $81$ .

$f'' (0) = \frac{9 \cdot 2}{9 \cdot 9}$

Cancel the common factor.

$f'' (0) = \frac{9 \cdot 2}{9 \cdot 9}$

Rewrite the expression.

$f'' (0) = \frac{2}{9}$

The final answer is $\frac{2}{9}$ .

$\frac{2}{9}$

The graph is concave up on the interval $(- 3, 3)$ because $f'' (0)$ is positive.

Concave up on $(- 3, 3)$ since $f'' (x)$ is positive

Substitute any number from the interval $(3, \infty)$ into the second derivative and evaluate to determine the concavity.

Replace the variable $x$ with $6$ in the expression.

$f'' (6) = \frac{- 2 {(6)}^{2} + 18}{{({(6)}^{2} + 9)}^{2}}$

Simplify the result.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$f'' (6) = \frac{- 2 \cdot 36 + 18}{{(6^{2} + 9)}^{2}}$

Multiply $- 2$ by $36$ .

$f'' (6) = \frac{- 72 + 18}{{(6^{2} + 9)}^{2}}$

Add $- 72$ and $18$ .

$f'' (6) = \frac{- 54}{{(6^{2} + 9)}^{2}}$

Simplify the denominator.

Raise $6$ to the power of $2$ .

$f'' (6) = \frac{- 54}{{(36 + 9)}^{2}}$

Add $36$ and $9$ .

$f'' (6) = \frac{- 54}{45^{2}}$

Raise $45$ to the power of $2$ .

$f'' (6) = \frac{- 54}{2025}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $- 54$ and $2025$ .

Factor $27$ out of $- 54$ .

$f'' (6) = \frac{27 (- 2)}{2025}$

Cancel the common factors.

Factor $27$ out of $2025$ .

$f'' (6) = \frac{27 \cdot - 2}{27 \cdot 75}$

Cancel the common factor.

$f'' (6) = \frac{27 \cdot - 2}{27 \cdot 75}$

Rewrite the expression.

$f'' (6) = \frac{- 2}{75}$

Move the negative in front of the fraction.

$f'' (6) = - \frac{2}{75}$

The final answer is $- \frac{2}{75}$ .

$- \frac{2}{75}$

The graph is concave down on the interval $(3, \infty)$ because $f'' (6)$ is negative.

Concave down on $(3, \infty)$ since $f'' (x)$ is negative

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Concave up on $(- 3, 3)$ since $f'' (x)$ is positive

Concave down on $(3, \infty)$ since $f'' (x)$ is negative

Find the Concavity natural log of x^2+9

Related Questions

Support LCMGCF.com by disabling your adblocker.